Question

A charge $q$ is placed at the centre of the line joining two equal charges $Q.$ The system of three charges will be in equilibrium, if $q$ is equal to

• A. $\frac{-Q}{2}$
• B. $\frac{-Q}{4}$
• C. $\frac{+Q}{2}$
• D. $\frac{+Q}{4}$

Answer 1

The correct option is B $\frac{-Q}{4}$

For system to be in equilibrium
$F_{BA}+F_{BC}=0$
$F_{BA}=-F_{BC}$
$\frac{K{Q}^2}{(2x{)}^2}=-\frac{KQq}{x^2}$
$q=-\frac{Q}{4}$